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| (401) Posted by Arno Tungler [Sunday, May 29, 2022 14:25] |
(398) is cooked with
1.c5 2.c4 3.c3 4.Qc7 5.Qc4 6.Qf4 7.Qxh6 8.Qg6 9.h6 10.Qh7 11.axb6 12.Sxa6 13.Sc7 14.Rxa5 15.Rxa4 16.Rg4 17.Rg6 18.f6 dxc7=
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| (402) Posted by Arno Tungler [Sunday, May 29, 2022 14:30] |
And similarly is (397)
1.c5 2.c4 3.c3 4.Qc7 5.Qf4 6.Qxh6 7.Qg6 8.h6 9.Qh7 10.axb6 11.Rxa4 12.Rg4 13.Rxg3 14.Rg6 15.f6 16.Sa6 17.Sc7 Bxc7=
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| (403) Posted by Arno Tungler [Sunday, May 29, 2022 14:49] |
Maybe something like this?.
 (= 15+16 )
Ser-h=19
1.axb6 2.Ra3 3.Rxd3 4.Rxd6 5.Rc6 6.d6 7.dxc5 8.Qd2 9.Sd7 10.Rd6 11.Rd4 12.Rxf4 13.Rxg4 14.Qxh6 15.Qg6 16.h6 17.Qh7 18.Rg6 19.f6 Bxc7=
Here also could wPa5 be added for a Ser-h=20.
EDITED because of cook…
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| (404) Posted by Arno Tungler [Sunday, May 29, 2022 18:36] |
Cooked with
1.axb6 2.cxd6 3.dxc5 4.Ra3 5.Rxd3 6.Rf3 7.Rxf4 8.Rxg4 9.Sa6 10.Qc7 11.Qd6 12.Sc7 13.Qd2 14.Qxh6 15.Qg6 16.h6 17.Qh7 18.Rg6 19.f6 Bxc7=
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| (405) Posted by Arno Tungler [Sunday, May 29, 2022 18:58] |
Another try…
 (= 15+16 )
Ser-h=19
1.axb6 2.Rxa3 3.Rxd3 4.Rd6 5.Rc6 6.d6 7.dxc5 8.Qd2 9.Sd7 10.Rd6 11.Rd4 12.Rxf4 13.Rxg4 14.Qxh6 15.Qg6 16.h6 17.Qh7 18.Rg6 19.f6 Bxc7=
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| (406) Posted by Michel Caillaud [Wednesday, Jun 8, 2022 15:39] |
(405) is cooked :
1.g×h6 2.Bg7 3.Kf8 4.f6 5.d6 6.d×ç5 7.Sç6 8.Qd7 9.Sd8 10.Sf7 11.Ké8 12.Kd8 13.Qé8 14.Qf8 15.a×b6 16.R×a3 17.R×d3 18.Rd7 19.Ké8 B×ç7=
Would be funny if the "missing" 32th piece had to be added (wRg6?)
I wondered about the use of wRg4 and found a cook with the move Rg3-g7 without it; so maybe wRg4 to g6 (??).
The (almost) same stalemate can be demonstrated with 16 pieces stalemated :
 (= 16+14 )
ser-!=22
1.ç×d3 2.Qç2 3.Qç3 4.d4 5.d3 6.Sd2 7.Qç6 8.Q×a4 9.Qd7 10.a4 11.a5 12.a×b6 13.Ra8 14.Ré8 15.R×é6 16.Rg6 17.Q×h3 18.Qg3 19.h3 20.Qh2 21.Rg3 22.f3!=
May seem off topic, but such a ser-!= can easily be turned in a ser-h= by reversing colors and adding a "technical" move. Not the more elegant way of producing a ser-h=..;
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| (407) Posted by Michel Caillaud [Wednesday, Jun 8, 2022 15:54] |
I agree that this kind of problem with ser-h= are interesting with the lowest number of moves.
Some other points of interest (for my taste) are keeping the 16 pieces stalemated (see 406) or using the lowest number of white pieces in the diagram position.
Here is an attempt for this last point :
 (= 7+16 )
ser-h=30
1.a×b6 2.Ra2 3.R×ç2 4.Rç6 5.Ré6 6.ç6 7.d5 8.d4 9.d3 10.d2 11.d1=R 12.Qd2 13.Qh6 14.Rd8 15.Sd7 16.ç5 17.ç4 18.ç3 19.ç2 20.ç1=B 21.Bg5 22.Bf6 23.g5 24.Qg7 25.h6 26.Rh7 27.Qh8 28.B6g7 29.Rg6 30.f6 é6=
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| (408) Posted by Olaf Jenkner [Wednesday, Jun 8, 2022 16:16] |
I'd be glad to see at least one correct setting of a ser-h=.
If there are some others we can look for the shortest ones or the ones with the lowest number of white pieces ...
But in this thread, as James created, we are looking for the longest problems per section.
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| (409) Posted by Arno Tungler [Wednesday, Jun 8, 2022 22:19] |
Great contributions by Michel for the ser-h= realm, I did not find quick cooks for the fantastic 30-mover with minimum white force AND the longest maneuver! Will still look for those… For my “shortest” version I agree to add the missing wRg6 - just to move g4 to g6 is not sufficient because of the dual 11.Rd3 12.Rg3 13. Qxf4 14.Qxh6 etc.
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| (410) Posted by Olaf Jenkner [Wednesday, Jun 8, 2022 22:36] |
I follow the thread every day and I am thrilled with your
problems. I did not add them to the list because the
soundness is very doubtful.
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| (411) Posted by Michel Caillaud [Friday, Jun 10, 2022 19:40] |
Some years ago, George Sphicas and I were faced with such a situation on the occasion of an article in StrateGems about records in captureless seriesmovers : some very nice (for our taste) problems were found but it was disheartening that ugly transformations (having King in check on diagram, adding uninteresting moves...) had to be made to produce records. So, though the article was about records, some problems were presented in two versions : the "artistic version" and the "record version".
BTW, afterwards, Arno cooked a number of problems and improved a number of records from this article, but that is another story...
What I guess Arno suggested (and I agree with him) is that making it the longest is not the more exciting for every stipulation.
For some stipulations, ingenuity of the composer is required, as is seen in a number of examples in this thread.
For some other stipulations (as ser-h=...), it can be a matter of adding some pieces to be captured to make it longer (of course, there is "technical work" for cooks to be avoided).
Here is an attempt for a "record version" from (407) :
 (= 13+16 )
ser-h=35
6 pieces are added to extend the record by 5 moves : 1.a×b6 2.R×a6 3.R×a5 4.Ra4 5.R×ç4 6.R×d4 7.R×d5 8.R×ç5 9.R×ç6 10.Ré6...
"Wonderful" will say the record-addict. "Horrible" will say the artist...
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| (412) Posted by Hauke Reddmann [Saturday, Jun 11, 2022 08:31] |
Now for something completely different: Did anyone sift
the (large) output of (say) k='homebase' and stip='ser' on PDB yet?
After all, the theme of this thread could have been
worked on before :-)
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| (413) Posted by James Malcom [Saturday, Jun 11, 2022 08:56] |
Hauke, with a few more keywords, all is easily pared down! k='homebase' and stip='ser' and position='sKe8 sDd8' AND NOT S='MatPlus.net Forum'
This gives some Bartel problems and more for a total of 12.
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| (414) Posted by Joost de Heer [Saturday, Jun 11, 2022 23:07] |
In Winchloe I found:
 (= 2+16 )
Hilmar Ebert, Alexander Lehmkuhl, feenschach 1976, ser-#11
So Arno's ser-# (post 39) is completely anticipated.
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| (415) Posted by Olaf Jenkner [Sunday, Jun 12, 2022 00:14] |
It's time for a 15th interim balance.
It's hard to find the current record in section Ser-s= . It might be (113).
Several attempts surpass his problem but the might be cooked as well.
(39) by Arno Tüngler is completely anticipated.
New section ser-h=. I didn't include ser-h= until now because I couldn't believe
it's possible to find a correct setting. Correctness is doubtful anyway ...
Current records are:
Ser-# ...11 ........... (39) Hilmar Ebert / Alexander Lehmkuhl (feenschach 1976)
Ser-= ...19 .......... (349) Arno Tüngler / Michael Schreckenbach
ser-s# ..17 .......... (124) Frank Richter
ser-s# ..17 .......... (128) Michael Schreckenbach
ser-s= ..22 .......... (113) Michel Caillaud after D. Stojnić and M. Mladenović (C?)
Ser-r# ..11 .......... (348) Michael Schreckenbach
Ser-r= ..17 .......... (273) Joost de Heer (C?)
Ser-!# ..14 .......... (325) Michael Schreckenbach / Frank Richter
Ser-!# MMM 16 .. (343) Michael Schreckenbach
Ser-!= ..26 .......... (118) Dragan Stojnić (C?)
Ser-h# .KNC 11 .. (42) Michel Caillaud / Arno Tüngler
Ser-h# .. 15 ........ (222) Arno Tüngler / Miodrag Mladenović
Ser-h= .. 35 ........ (411) Michel Caillaud (C?)
Ser-hs# .13 ........ (331) Dragan Stojnić / Miodrag Mladenović (C?)
Ser-hs= .18 ........ (202) Miodrag Mladenović / Arno Tüngler (C?)
Everybody is invited to surpass one of the entries or to cook one marked C?.
All entries of the record list and (131) can be seen and autoplayed in the PDB:
https://pdb.dieschwalbe.de/search.jsp?expression=probid%3E%27P1400890%27and+probid%3C%27P1403155%27and+CREATIONUSER+%3D%27jenkner%27%0D%0A
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| (416) Posted by Michel Caillaud [Sunday, Jun 12, 2022 12:17] |
There are some attempts ("not yet cooked" ?) missing in the list :
ser-s= : (389); Arno came close to a cook, but not quite : (392)
ser-r= : (396)
ser-!= : (386)
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| (417) Posted by Arno Tungler [Friday, Jun 24, 2022 23:01] |
(202) ser-hs=18 is cooked with
1.Sa6 2.b5 3.b4 4.bxc3 5.Rb8 6.Rb5 7.Rxf5 8.Rxg5 9.Rxg2 10.Rxf2 11.Rxe2 12.Rxh2 13.Rxh4 14.Rxd4 15.Rxd6 16.Rb6 17.Rxb3 18.Rxb2 and Bxf7+ Kxf7=
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| (418) Posted by Arno Tungler [Saturday, Jun 25, 2022 12:01] |
In the (411) Ser-h=35 Michel should at least change the places of Ba5 and Sc5 due to the cook
1.Sxc6 2.Sb4 3.Sxa6 4.c6 5.Sc7 6.cxd5 7.dxc4 8.c3 9.c2 10.c1=B 11.Bb2 12.axb6 13.Rxa5 14.Ra4 15.Rxd4 16.Rd6 17.Bxe5 18.g5 19.Rg6 20.Beg7 21.Bh6 22.f6 23.Kf7 24.Bfg7 25.Qf8 26.Ke8 27.Kd8 28.Qe8 29.Bf8 30.Rg7 31.Rf7 32.Bfg7 33.Qf8 34.Ke8 35.Se6 Sxe6=
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| (419) Posted by Arno Tungler [Saturday, Jun 25, 2022 15:06] |
(411) is also cooked with
1.axb6 2.Rxa6 3.Rxa5 4.Ra4 5.Rxc4 6.Rxc5 7.Rxc6 8.Re6 9.c5 10.c4 11.c3 12.c2 13.c1=B 14.Bh6 15.g5 16.Sf6 17.Bfg7 18.O-O 19.d6 20.Qd7 21.Rd8 22.Qe8 23.Kf8 24.Sg8 25.Sd7 26.Rg6 27.f6 28.Kf7 29.Qf8 30.Ke8 31.Bh8 32.Rg7 33.Rf7 34.B6g7 35.h6 e6=
With castling in the solution!
That means that the switch of the two white units is not sufficient. Will also see whether something similar is possible for the shorter version…
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| (420) Posted by Michel Caillaud [Tuesday, Aug 2, 2022 00:50] |
I didn't notice until now that Arno did cook (411).
Congratulations again, Arno!
I suppose I should try to correct...
Maybe like this :
 (= 13+16 )
ser-h=35
same solution
as Arno indicated, not having a Knight on c5, is sufficient to avoid the first cook.
Queen d5 is aimed to prevent black King to use square f7 at a critical moment and so to prevent the second cook(?!).
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